Let $R$ be a local Noetherian ring and $\mathfrak{m}$ its maximal ideal. I have shown that either $\mathfrak{m}^n\neq\mathfrak{m}^{n+1}$ or $\mathfrak{m}^n = 0$ and supposedly in this latter case $R$ is Artinian. But I am having a hard time proving this.
The easiest way seemed to me to create a composition series $R\supsetneq\mathfrak{m}\supsetneq\mathfrak{m}^2\supsetneq\cdots\supsetneq\mathfrak{m}^{n+1}=0$ since an $R$-module is Noetherian and Artinian iff it has a composition series. I can show $R/\mathfrak{m}$ is simple but not the other quotients. I am not really sure if they have to be simple. I also tried to prove they are isomorphic with $R/\mathfrak{m}$ but I struggle to prove the injectivity of such a morphism. Is the way I am trying to prove this possible or do I need to use something else? (e.g. this)
Best Answer
Here's a finer statement:
Without Noetherian assumptions, the implication 1$\Rightarrow$2 is strict [ref] and also 2$\Rightarrow$4 [ref].
Proof. 1$\Rightarrow$2. In an Artin ring the nilradical equals the Jacobson radical [AM, 8.2]. Since $A$ is Noetherian [AM, 8.5], the nilradical is nilpotent.
2$\Rightarrow$3. Let $\mathfrak{m}\subset A$ be a nilpotent maximal ideal and let $I\subset A$ be a radical proper ideal. Then $\mathfrak{m}^r\subset I$ for some integer $r\geq 1$, whence $\mathfrak{m}=\sqrt{\mathfrak{m}^r}\subset \sqrt{I}=I$. Thus $\mathfrak{m}=I$ by maximality of $\mathfrak{m}$.
3$\Rightarrow$4. Every prime ideal is proper radical.
4$\Leftrightarrow$5. The nilradical equals the intersection of all prime ideals.
4$\Rightarrow$3. A radical ideal equals the intersection of the prime ideals containing it.
Now suppose $A$ is Noetherian.
4$\Rightarrow$1. $A$ is local. On the other hand, it is of dimension $0$ and Noetherian, and thus, Artin [AM, 8.5]. $\square$
References
AM. Atiyah, Macdonald, Introduction to Commutative Algebra.